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1Bch21

# 1Bch21 - Review information for the indicated chapter.

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Chapter 21 Here are a few sample problems illustrating how to find electric fields using Coulomb’s law and superposition. First, a brief side-note. For several parts of your homework, you need to be able to Taylor expand an expression of the form ( β + α ) - p in the limit β >> α . This expansion is as follows: ( β + α ) - p = β - p 1 + α β - p β - p " 1 - p α β + ( - p )( - p - 1) 2 α β 2 + .... # This expansion is valid only if β >> α , so that α/β << 1. As α/β increases, more terms of the above expansion need to be included in order to accurately approximate ( β + α ) - p . In most cases in your homework, you only need the first two terms of this expansion. The exception is the end of question 21.95, which says to show that a certain force is proportional to x - 3 . In this case, you need the first three terms (shown above). The use of this expansion is illustrated in the sample problems below. The primary purpose of these problems is to illustrate how to find electric fields using Coulomb’s law and superposition. Also recall that in this context, k = 1 4 π² 0 Problem 1 (medium): Three point charges are placed on the y -axis. A charge Q 1 is placed at (0 , b ), a charge Q 2 is placed at (0 , 0), and a charge Q 3 is placed at (0 , - b ). 1: What is the electric field at a point ( x, 0), where x 6 = 0? 2: For distant points ( x, 0) where x >> b , find approximate expressions for the electric field components. That is, write the components in the form E x ( x, 0) A 1 x x 2 + A 2 x x 4 E y ( x, 0) B 1 y x 3 + B 2 y x 5 and find A 1 x , A 2 x , B 1 y , and B 2 y . Solution: Suppose we have point charges { Q 1 , Q 2 , ..., Q N } , where each charge Q j is lo- cated at a point ( x j , y j ). We wish to evaluate the electric field due to these charges at a point ( x, y ). For charge Q j , let r j denote the distance from ( x j , y j ) to ( x, y ). Let θ j denote the angle, to the positive x -direction, of the vector r j starting at ( x j , y j ) and ending at ( x, y ). Then the components of the electric field at ( x, y ) are as follows: 1

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E x ( x, y ) = N X j =1 kQ j r 2 j cos θ j = N X j =1 kQ j r 2 j x - x j r j = N X j =1 kQ j ( x - x j ) r 3 j = N X j =1 kQ j ( x - x j ) [( x - x j ) 2 + ( y - y j ) 2 ] 3 / 2 E y ( x, y ) = N X j =1 kQ j r 2 j sin θ j = N X j =1 kQ j r 2 j y - y j r j = N X j =1 kQ j ( y - y j ) r 3 j = N X j =1 kQ j ( y - y j ) [( x - x j ) 2 + ( y - y j ) 2 ] 3 / 2 In this problem, we have three charges { Q 1 , Q 2 , Q 3 } whose locations are ( x 1 , y 1 ) = (0 , b ) ( x 2 , y 2 ) = (0 , 0) ( x 3 , y 3 ) = (0 , - b ) We wish to evaluate the electric field at a point ( x, 0) along the x -axis. To do so, we simply insert our values into the above expressions for E x ( x, y ) and E y ( x, y ): E x ( x, 0) = kQ 1 ( x - 0) [( x - 0) 2 + (0 - b ) 2 ] 3 / 2 + kQ 2 ( x - 0) [( x - 0) 2 + (0 - 0) 2 ] 3 / 2 + kQ 3 ( x - 0) [( x - 0) 2 + (0 + b ) 2 ] 3 / 2 = k ( Q 3 + Q 1 ) x ( x 2 + b 2 ) 3 / 2 + kQ 2 x 2 E y ( x, 0) = kQ 1 (0 - b ) [( x - 0) 2 + (0 - b ) 2 ] 3 / 2 + kQ 2 (0 - 0) [( x - 0) 2 + (0 - 0) 2 ] 3 / 2 + kQ 3 (0 + b ) [( x - 0) 2 + (0 + b ) 2 ] 3 / 2 = k ( Q 3 - Q 1 ) b ( x 2 + b 2 ) 3 / 2 We now consider part 2 . For points ( x, 0) where
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1Bch21 - Review information for the indicated chapter.

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